# How to calculate the intеgral?

I need to calculate this definite integral: $$\int_\frac{\pi}{6}^\frac{5\pi}{6}\sqrt{(\sin(t)\cos(t))^2}\,dt$$ I can`t uncover this root, because of interval of x. Then what should I do? Help please.

• One may recall that $\sqrt{x^2}=|x|$. – Olivier Oloa Dec 15 '17 at 19:40
• $$\frac{1}{2}\int_{\pi/6}^{5\pi/6}\left|\sin(2t)\right|\,dt=\frac{1}{4}\int_{\pi/3}^{5\pi/3}\left|\sin\theta\right|d\theta$$ is not that difficult to compute. – Jack D'Aurizio Dec 15 '17 at 19:41

$$\int_\frac{\pi}{6}^\frac{5\pi}{6}\sqrt{(\sin(t)\cos(t))^2}\,dt= \int_\frac{\pi}{6}^\frac{5\pi}{6}|\sin(2t)|/2\,dt= {1\over 4}\int_\frac{\pi}{3}^\frac{5\pi}{3}|\sin(x)|\,dx= {1\over 4}\int_\frac{\pi}{3}^\frac{5\pi}{3}\sin(x)\,dx$$